Yes, giving an empty list of variables is the same as giving no
variables at all, and no, Solve does not assume that all symbols are
variables. It assumes that y and b are variables and a and x parameters.
The reason why it makes these choices seems a little subtle.
The clearest indication that Solve is not assuming that all symbols are
variables is the fact that it does not produce the message that it
produces here:
Solve[{b*x + y == 7 && a*x - y == 1}, {x, y, a, b}]
Solve::svars:Equations may not give solutions for all "solve" variables.
>>
{{y -> -1 + a*x, b -> (8 - a*x)/x}}
In fact, what Solve seems to do is something like the following.
First y is expressed in terms of x and b as y = 7 - b x. The reason
why Solve choses to treat y in this way (as a "variable" rather than a
"parameter") by doing so it does not have to make any assumptions at all
about the other symbols. If it chose x as a "variable" (meaning, that it
would have tried to express x in terms of y and a) it would have to make
the assumption that a is non-zero and it wants to make as few
assumptions as possible. Similarly, if it chose a as a variable, it
would have to assume that x is non-zero. The choice of you is the only
one that allows it to make no assumptions at this stage.
However, as a result of this choice, Solve is now "forced" to treat a
and x as "parameters" (since they were used in the solution for y).
Now, by substituting for y into the second equation, Solve gets an
equation involving a, b, and x. However, since a and x are now being
treated as "parameters", b has to became a "variable" and the solution
obtained turns out to be exactly the same as given by:
Solve[{x + b*y == 7 && a*x - y == 1}, {y, b}]
{{y -> -1 + a*x, b -> (7 - x)/(-1 + a*x)}}
You can easily check that this behaviour is not dependent on the symbols
used for the variables - only on their role in the equations.
Next, something that seems to me more interesting. If you change these
equations somewhat you will see behaviour that may appear even more
surprising.
Solve[{x + b*y == 7 && a*x - y == 1}]
{{a -> (1 + y)/x, b -> (7 - x)/y}, {a -> 1/7, y -> 0, x -> 7}, {b -> -7, y -> -1, x -> 0}}
Here Solve returned a list of solutions rather than one solution. The
first is the same one it returns when called with variable specification
{a,b}:
Solve[{x + b*y == 7 && a*x - y == 1}, {a, b}]
{{a -> (1 + y)/x, b -> (7 - x)/y}}
This solution of course requires assumptions on the "parameters" x,y
i.e. x!=0 and y!=0. In general, when called with specific variables,
Solve by default does not return solutions that involve assumptions on
parameters. However the assumption that a parameter is non-zero is
normally not excluded, in other words, it is not really considered "an
assumption". But when called without specified variables Solve actually
does something that it normally does not do : it actually considers the
special cases x==0 and y==0 separately and returns two
additional solutions.
Note how similar the result of Solve is to the one given by Reduce:
Solve[{x + b*y == 7 && a*x - y == 1}]
{{a -> (1 + y)/x, b -> (7 - x)/y}, {a -> 1/7, y -> 0, x -> 7}, {b -> -7, y -> -1, x -> 0}}
Reduce[{x + b*y == 7 && a*x - y == 1}]
(y == -1 && x == 0 && b == -7) || (y == 0 && x == 7 =
&& a == 1/7) ||
(y != 0 && b == (7 - x)/y && x != 0 && a == (1 + y)/x)
All this is completely undocumented and in fact even appears to
contradict the documentation, since Solve does appear to be actually
returning solutions that require parameters to have special values. Of
course only stated that this is not done when variables are explicitly
specified. Since we do not specify variables we are in undocumented
territory and Solve is not obliged to follow the behavior documented for
the case of specified variables.
Andrzej Kozlowski
On 5 Apr 2013, at 11:33, debguy <johnandsara2 at cox.net> wrote:
> test if these are the same (all Symbols assumed variables, i think):
>
> Solve[a x + y == 7 && b x - y == 1, {}]
>
> Solve[a x + y == 7 && b x - y == 1]
> On 4 Apr 2013, at 11:53, juchheissassa at gmail.com wrote:
>
>> I happened to hit on the following curious thing:
>>
>> If I use Solve with {} instead of vars, I actually get a solution.
>>
>> Is this documented anywhere? What are the rules?
>>
>> e.g.:
>>
>> Solve[a x + y == 7 && b x - y == 1, {}] yields {{y -> 7 - a x, b -> (8 - a x)/x}}
>>
>> while
>>
>> Solve[a x + y == 7 && b x - y == 1, {x,y}] yields {{x -> 8/(a + b), y -> -((a - 7 b)/(a + b))}}
>>
>
>>
>